Câu 1: \(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{6}{19}\)
\(\Leftrightarrow\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{18}{19}\)
\(\Leftrightarrow1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{18}{19}\)
\(\Leftrightarrow1:\left(x+3\right)=\dfrac{1}{19}\)
=>x+3=19
hay x=16